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Let

  • $T>0$
  • $\lambda^1$ denote the Lebesgue measure on $\mathbb R$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\in[0,\:T]}$ be a filtration on $(\Omega,\mathcal A)$
  • $(B)_{t\in[0,\:T]}$ be a real-valued $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
  • $(\Phi)_{t\in[0,\:T]}$ be a real-valued $\mathcal F$-progressively measurable stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $$\Phi(\omega)\in\mathcal L^2\left(\left.\lambda^1\right|_{[0,\:T]}\right)\;\;\;\text{for }\operatorname P\text{-almost all }\omega\in\Omega\tag 1$$
  • $0=t_0<\cdots<t_N=T$ for some $N\in\mathbb N$, $\varsigma:=\left\{t_0,\ldots,t_N\right\}$, $$\left|\varsigma\right|:=\max_{1\:\le\:i\:\le\:N}\left(t_i-t_{i-1}\right)$$ and $$\Phi_t^\varsigma:=\sum_{i=1}^N1_{\left(t_{i-1},\:t_i\right]}(t)\Phi_{t_{i-1}}\;\;\;\text{for }t\in[0,T]\tag 0$$

How can we show that, if $\Phi$ is almost surely continuous, $$\sup_{t\in[0,\:T]}\left|\int_0^t\Phi_s^\varsigma-\Phi_s\:{\rm d}B_s\right|\xrightarrow{\left|\varsigma\right|\to0+}0\tag 2$$ in probability?

It's clear that $$\Phi^\varsigma\xrightarrow{\left|\varsigma\right|\to0+}\Phi\tag 3$$ uniformly almost surely and hence $$\int_0^T\Phi_t^\varsigma\:{\rm d}t\xrightarrow{\left|\varsigma\right|\to0+}\int_0^T\Phi_t\:{\rm d}t\tag 4$$ almost surely. Note also that $$\int_0^t\Phi_s\:{\rm d}B_s=\sum_{i=1}^N\Phi_{t_{i-1}}\left(B_{t_i\:\wedge\:t}-B_{t_{i-1}\:\wedge\:t}\right)\;\;\;\text{for all }t\in[0,T]\:.\tag 5$$

For those $\Phi$, which even satisfy $$\left\|\Phi\right\|:=\operatorname E\left[\int_0^T\Phi_t^2\:{\rm d}t\right]<\infty\;,\tag 6$$ I've defined $$\Phi\cdot B:=\left(\int_0^t\Phi_s\:{\rm d}B_s\right)_{t\in[0,\:T]}$$ to be the unique (up to indistinguishability) almost surely continuous $\mathcal F$-martingale with $$\left\|\Phi^n\cdot B-\Phi\cdot B\right\|_{\mathcal M^2}:=\left\|(\Phi^n\cdot B)_T-(\Phi\cdot B)_T\right\|_{\mathcal L^2(\operatorname P)}\xrightarrow{n\to\infty}0\tag 7$$ for some "elementary" processes (i.e. processes of the form $(0)$ whose Itō integral process is defined by $(5)$) $\Phi^n$.

For the more general $\Phi$, $\Phi\cdot B$ is defined by "localization" as usual.

So, two Itō integral process coincide, if they coincide with respect to $\left\|\;\cdot\;\right\|_{\mathcal M^2}$. With that in mind, I have no clue how I can show $(2)$, since I don't see a way to use the Itō isometry $$\left\|\Phi\cdot B\right\|_{\mathcal M^2}=\left\|\Phi\right\|\;.\tag 8$$

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  • $\begingroup$ There appears to be a typo in (2). $\endgroup$ Commented Nov 5, 2016 at 16:41
  • $\begingroup$ @JohnDawkins Fixed. Thank you. $\endgroup$
    – 0xbadf00d
    Commented Nov 5, 2016 at 16:44
  • $\begingroup$ I might miss the point of your question. But in the case that $\Phi$ is square integrable w.r.t time and p. measure, it seems to me that you already have convergence of $(2)$ in $L^2(\Omega)$. So in the general case you take a localising sequence. Then you get: $P( (2) < \epsilon) \le P( sup_{t \le \tau_n} |\text{real martingale}| < \epsilon ) + P(t \ge \tau_n)$. Now both the latter terms are small. $\endgroup$
    – Kore-N
    Commented Nov 5, 2016 at 17:39
  • $\begingroup$ @Cornelis I don't want to assume that $Φ\in\mathcal L^2\left(\operatorname P\otimes\left.λ^1\right|_{[0,\:T]}\right)$, but the weaker assumption on $Φ$ stated in the question (there was a typo in a former version). However, even with the stronger assumption, $$\int_0^TΦ_t^\varsigma\:{\rm d}B_t\xrightarrow{|\varsigma|\to0+}\int_0^TΦ_t\:{\rm d}B_t$$ won't hold in $L^2(\operatorname P)$ unless, for example, $Φ$ is continuous in mean-square, which is not assumed here and isn't implied by the assumptions. $\endgroup$
    – 0xbadf00d
    Commented Nov 5, 2016 at 18:46
  • $\begingroup$ I am still not sure that we are on the same page. I still think that a "usual localisation" works. For example if you choose $\tau_n = inf \{ t : |\Phi_t| \ge n$ or $B_t \ge n\}$ your issue does not show up: you get convergence in $L^p(\Omega)$ for any $p$ through dominated convergence. Of course here you need to use continuity. $\endgroup$
    – Kore-N
    Commented Nov 5, 2016 at 19:04

1 Answer 1

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Consider $\Phi^{\zeta}$ and $\Phi$ as in the question. From the continuity it follows that $\Phi^{\zeta}$ converges pointwise to $\Phi$.

Consider the stopping times $\tau_n = T \wedge \inf \{t \ge 0 : |\Phi_t| \ge n \text{ or } |B_t| \ge n\}.$ This sequence is localizing for the Brownian motion on $[0,T]$.

Now $P( \sup_{t\in[0,\:T]}\left|\int_0^t\Phi_s^\zeta-\Phi_s\:{\rm d}B_s\right| < \epsilon) \le P( \sup_{t\in[0,\tau_n]}\left|\int_0^t\Phi_s^\zeta-\Phi_s\:{\rm d}B_s\right| < \epsilon) + P(t \ge \tau_n).$ The latter term is small by monotone convergence. Let us concentrate on the first term. $$\sup_{t\in[0,\tau_n]}\left|\int_0^t\Phi_s^\zeta-\Phi_s\:{\rm d}B_s\right| = \sup_{t\in[0, T]}\left|\int_0^t\Phi_{s \wedge \tau_n}^\zeta-\Phi_{s \wedge \tau_n}\:{\rm d}B_{s \wedge \tau_n}\right|$$

Now what we have written on the RHS, inside the supremum is a real martingale, say $X_t$. So we can just apply Doob's maximal inequality: $P(sup_{t \in [0, T]} |X_t| \ge \epsilon) \le C \frac{\mathbb{E}[X_T^2]}{\epsilon^2}$

After this step the most is done: we have reduced ourselves to looking at the square mean of a stochastic integral. By localisation we know that we can apply Ito isometry. Hence: $$\mathbb{E}[X_T^2] = \int_0^T(\Phi_{s \wedge \tau_n}^\zeta-\Phi_{s \wedge \tau_n})^2 ds$$

Now by dominated convergence (note that we have localized in such a manner that the $\Phi's$ are bounded) this expression goes to zero over $\zeta.$

Now we can conclude, by choosing first $n$ and then $\zeta$ large enough.

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  • $\begingroup$ I will check your proof soon. Meanwhile, I've got a very elegant idea for a proof an even more general version of the statement. However, I'm struggling with an apparently easy statement. Maybe you can help. $\endgroup$
    – 0xbadf00d
    Commented Nov 6, 2016 at 0:15

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